Question

integrate

Answer 1

integrate?

Answer 2

if we are integrating nothing, the integral of 0 is 0 haha. Are you missing a function we have to integrate or what?

Answer 3

\[\int\limits_{0}^{1} x^2-x/(x^2-1)(x+1)\]

Answer 4

LOL. Kids had to stop for a few min

Answer 5

First of all simplify the expression algebraically so that integration will be simpler.
x^2 - x/(x^2 - 1)(x + 1) = x(x-1)/(x+1)(x-1)(x+1)
x(x+1)^2

Answer 6

\[\int\limits_{0}^{1} \frac{x}{(x+1)^{2}} dx = \int\limits_{0}^{1} \frac{x}{x^2 + 2x + 1} dx\]

Answer 7

Got you which equals \[\ln (2)-\pi/4\approx0.092251\]

Answer 8

\[\int\limits_{0}^{1} \frac{x}{(x+1)^{2}} dx =\ln(2)-\frac{1}{2}\]

Answer 9

Did you let u = x + 1 so that du = dx ; x = u - 1\[\int\limits_{1}^{2} \frac{u-1}{(u)^{2}} du\]
For the limits when x = 0, u = 1 and when x = 1 , u = 2

Answer 10

\[\int\limits_{1}^{2} \frac{1}{u} - \frac{1}{u^{2}} du\]

Answer 11

ln|u| + 1/u: Evaluated from 1 to 2
(ln(2) + 1/2) - (ln(1) + 1)..........remember ln(1) = 0
ln(2) + 1/2 - 1 = ln(2) + 0.5 = 1.193147

Answer 12

the answer is \[\ln(2)-\frac{1}{2}\approx .19314718056\]

Answer 13

yes zarkon is write I overlooked that.